/** * Compares two strings lexicographically. * The comparison is based on the Unicode value of each character in * the strings. The character sequence represented by this * <code>String</code> object is compared lexicographically to the * character sequence represented by the argument string. The result is * a negative integer if this <code>String</code> object * lexicographically precedes the argument string. The result is a * positive integer if this <code>String</code> object lexicographically * follows the argument string. The result is zero if the strings * are equal; <code>compareTo</code> returns <code>0</code> exactly when * the {@link #equals(Object)} method would return <code>true</code>. * <p> * This is the definition of lexicographic ordering. If two strings are * different, then either they have different characters at some index * that is a valid index for both strings, or their lengths are different, * or both. If they have different characters at one or more index * positions, let <i>k</i> be the smallest such index; then the string * whose character at position <i>k</i> has the smaller value, as * determined by using the < operator, lexicographically precedes the * other string. In this case, <code>compareTo</code> returns the * difference of the two character values at position <code>k</code> in * the two string -- that is, the value: * <blockquote><pre> * this.charAt(k)-anotherString.charAt(k) * </pre></blockquote> * If there is no index position at which they differ, then the shorter * string lexicographically precedes the longer string. In this case, * <code>compareTo</code> returns the difference of the lengths of the * strings -- that is, the value: * <blockquote><pre> * this.length()-anotherString.length() * </pre></blockquote> * * @param anotherString the <code>String</code> to be compared. * @return the value <code>0</code> if the argument string is equal to * this string; a value less than <code>0</code> if this string * is lexicographically less than the string argument; and a * value greater than <code>0</code> if this string is * lexicographically greater than the string argument. */ public int compareTo(String anotherString) { int len1 = count; int len2 = anotherString.count; int n = Math.min(len1, len2); char v1[] = value; char v2[] = anotherString.value; int i = offset; int j = anotherString.offset; if (i == j) { int k = i; int lim = n + i; while (k < lim) { char c1 = v1[k]; char c2 = v2[k]; if (c1 != c2) { return c1 - c2; } k++; } } else { while (n-- != 0) { char c1 = v1[i++]; char c2 = v2[j++]; if (c1 != c2) { return c1 - c2; } } } return len1 - len2; } 要明白就自己看看源代码
分别表示小于 等于 大于。
* Compares two strings lexicographically.
* The comparison is based on the Unicode value of each character in
* the strings. The character sequence represented by this
* <code>String</code> object is compared lexicographically to the
* character sequence represented by the argument string. The result is
* a negative integer if this <code>String</code> object
* lexicographically precedes the argument string. The result is a
* positive integer if this <code>String</code> object lexicographically
* follows the argument string. The result is zero if the strings
* are equal; <code>compareTo</code> returns <code>0</code> exactly when
* the {@link #equals(Object)} method would return <code>true</code>.
* <p>
* This is the definition of lexicographic ordering. If two strings are
* different, then either they have different characters at some index
* that is a valid index for both strings, or their lengths are different,
* or both. If they have different characters at one or more index
* positions, let <i>k</i> be the smallest such index; then the string
* whose character at position <i>k</i> has the smaller value, as
* determined by using the < operator, lexicographically precedes the
* other string. In this case, <code>compareTo</code> returns the
* difference of the two character values at position <code>k</code> in
* the two string -- that is, the value:
* <blockquote><pre>
* this.charAt(k)-anotherString.charAt(k)
* </pre></blockquote>
* If there is no index position at which they differ, then the shorter
* string lexicographically precedes the longer string. In this case,
* <code>compareTo</code> returns the difference of the lengths of the
* strings -- that is, the value:
* <blockquote><pre>
* this.length()-anotherString.length()
* </pre></blockquote>
*
* @param anotherString the <code>String</code> to be compared.
* @return the value <code>0</code> if the argument string is equal to
* this string; a value less than <code>0</code> if this string
* is lexicographically less than the string argument; and a
* value greater than <code>0</code> if this string is
* lexicographically greater than the string argument.
*/
public int compareTo(String anotherString) {
int len1 = count;
int len2 = anotherString.count;
int n = Math.min(len1, len2);
char v1[] = value;
char v2[] = anotherString.value;
int i = offset;
int j = anotherString.offset; if (i == j) {
int k = i;
int lim = n + i;
while (k < lim) {
char c1 = v1[k];
char c2 = v2[k];
if (c1 != c2) {
return c1 - c2;
}
k++;
}
} else {
while (n-- != 0) {
char c1 = v1[i++];
char c2 = v2[j++];
if (c1 != c2) {
return c1 - c2;
}
}
}
return len1 - len2;
}
要明白就自己看看源代码
public static void main(String[] args) {
String s = "abc";
String s1 = "bca";
String s2 = "cab";
String s3 = "abc";
System.out.println(s.compareTo(s1));
System.out.println(s.compareTo(s2));
System.out.println(s.compareTo(s3));
System.out.println(s2.compareTo(s1));
}
}
打印结果分别是:
-1
-2
0
1
String str1 = "abcdefg";
String str2 = "gfedcba";
public static void main(String args[]){
JavaTest javaTest = new JavaTest();
int flag = javaTest.str1.compareTo(javaTest.str2);
if(flag==0){
System.out.println("str1 = str2"+"----str1.compareTo(str2) = "+flag);
}else if(flag>0){
System.out.println("str1 > str2"+"----str1.compareTo(str2) = "+flag);
}else{
System.out.println("str1 < str2"+"----str1.compareTo(str2) = "+flag);//被执行,打印str1 < str2----str1.compareTo(str2) = -6
}
}}
结论,字母的先后顺序和大小写决定字母的大小